Question

Consider a reversible heat engine that employs a hot reservoir at a temperature of 610 K and a cold reservoir at 290 K. (a) What is the entropy change of the hot reservoir during a period in which 4730 J is extracted from the hot reservoir? 7.75 J/K

Answer 1

To solve this problem it is necessary that we start from the definition of entropy as a function of heat and temperature exchange. Mathematically this thermodynamic expression can be described as

`\Delta S = (Q)/(T)`

Where,

Q= Heat exchange

T = Temperature

Since we look for entropy in the hot reservoir, and considering our given values we have to

`T_H = 610K\\T_C = 290K \\Q_H = 4730J`

Replacing we have:

`\Delta S_H= (Q_H)/(T_H)`

`\Delta S_H = (4730J)/(610K)`

`\Delta S_H = 7.754J/K`

Therefore the final change in the entropy is 7.75J/K